If y-log e log e x - then dy dx is equal to

The correct option is A 1 x (log _e2 )^2 logx dydx = 1(log_2^x).log_e^x.1x.log^x_e = 1x.(log_e^2)^x.logx #160; ...

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Algebra of Derivatives

If y=log e ...

Question

If $y = {log}_{e} ({log}_{e} x)$ , then $\frac{d y}{d x}$ is equal to

\frac{1}{x (l o g_{e} 2)^{2} l o g x}

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\frac{1}{x {log}_{e} x {log}_{e} 2}

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\frac{1}{{log}_{e} {(2 x)}^{x}}

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\frac{1}{x log x {({log}_{e} 2)}^{2}}

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1 + \frac{x {log}_{e} 2}{1!} + \frac{x^{2}}{2!} ({log}_{e} 2)^{2} + \frac{x^{3}}{3!} ({log}_{e} 2)^{3} + . . . . . . \infty

{lim}_{x \to 1} \frac{x^{2} + x {log}_{e} x - {log}_{e} x - 1}{(x^{2} - 1)}

\frac{d y}{d x} = (x - y)^{2}

y (1) = 1

y = \tan^{- 1} \{\frac{\log_{e} (e / x^{2})}{\log_{e} (e x^{2})}\} + \tan^{- 1} (\frac{3 + 2 \log_{e} x}{1 - 6 \log_{e} x})

\frac{d^{2} y}{d x^{2}} =

y = {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{log (\frac{e}{x^{3}})}{log ({e x}^{3})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

\frac{d y}{d x} = ?

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